Subject: cube roots

Hi. My name is Heather Jones. I'm in 11 grade. What do the graphs of cube roots look like?

Hi Heather,

There are (at least) a couple of ways to 'see' this.

  1. Visual. Draw a graph of what y = x3 (the cube) looks like. Now YOU want the graph where x and y switch roles: y3 = x or y = x1/3. (This, by the way, is another standard notation for cube root. It is the notation you will need if you want to calculate cube roots of specific numbers on a calculator.)

    You can get this 'reversed' graph by taking the first one (drawn with the same scale on the x and y axes) and putting down a mirror on y = x. Now reflect the first graph in this mirror. Y becomes x, x becomes y and the first graph y=x3 becomes y3=x.

    Another way to see this is to draw the graph of y = x3 on a transparency that you would use in an overhead projector. Now switch the roles of x and y by labeling the horizontal axis y, and the vertical axis x. Finally turn the transparency over and rotate it so that the axis labeled y is vertical, with positive upwards, and the axis labeled x is horizontal with positive to the right. What you see is the graph of y = x1/3.
    Harley

    This is a general technique for turning the graph of ANY function into the graph of the 'inverse' function. Works for y=x2 to y2=x (or equivalently y=x1/2 - the square root).

    Just as the graph of y=x3 goes up far faster than y=x2 or y=x, the graph of y=x1/3 grows SLOWER than the graph of y=x1/2 which, in turn, is slower than y=x.

  2. Numerical. Take pairs of points from y=x3. (-2,-8), (-1,-1), (0,0), (1,1), (2,8) etc. Reverse each of the pairs: (-8,-2), (-1,-1) (0,0) (1,1) (8,2) etc. These reversed points (the mirror images in 1.) are points on the graph: cube root of -8 is -2, etc. Use these points (and others as necessary) to plot the graph.

    (3) With a calculator, you can get points on the graph just by plugging in points, and knowing that cube root of x is the same as x to the exponent (1/3). Again, enough points will give you information for a sketch of the graph.

Cheers,
Walter
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